Answer
$$\frac{{{x^2}}}{4} + \frac{1}{2}\ln \left| x \right| - \frac{1}{{{2^x}\ln 2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {\frac{1}{2}x + \frac{1}{{2x}} + \frac{1}{{{2^x}}}} \right)} dx \cr
& {\text{sum rule for derivatives}} \cr
& = \int {\frac{1}{2}x} dx + \int {\frac{1}{{2x}}} + \int {\frac{1}{{{2^x}}}} dx \cr
& {\text{use the constant multiple rule }}\int {kf\left( x \right)dx} = k\int {f\left( x \right)} dx \cr
& = \frac{1}{2}\int x dx + \frac{1}{2}\int {\frac{1}{x}} + \int {\frac{1}{{{2^x}}}} dx \cr
& {\text{rewrite the last integral}} \cr
& = \frac{1}{2}\int x dx + \frac{1}{2}\int {\frac{1}{x}} - \int {{2^{ - x}}} \left( { - 1} \right)dx \cr
& {\text{Integrate using the rules of integration}} \cr
& \int {{x^r}dx} = \frac{{{x^{r + 1}}}}{{r + 1}} + C,\,\,\,\,\int {\frac{1}{x}} dx = \ln \left| x \right| + C{\text{ }}\,\,{\text{and }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C,\,\,\, \cr
& \cr
& = \frac{1}{2}\left( {\frac{{{x^2}}}{2}} \right) + \frac{1}{2}\ln \left| x \right| - \frac{{{2^{ - x}}}}{{\ln 2}} + C \cr
& {\text{simplifying}} \cr
& = \frac{{{x^2}}}{4} + \frac{1}{2}\ln \left| x \right| - \frac{1}{{{2^x}\ln 2}} + C \cr} $$
